Milking Cows
Three farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins at time 700 and ends at time 1200. The third farmer begins at time 1500 and ends at time 2100. The longest continuous time during which at least one farmer was milking a cow was 900 seconds (from 300 to 1200). The longest time no milking was done, between the beginning and the ending of all milking, was 300 seconds (1500 minus 1200).
Your job is to write a program that will examine a list of beginning and ending times for N (1 <= N <= 5000) farmers milking N cows and compute (in seconds):
• The longest time interval at least one cow was milked.
• The longest time interval (after milking starts) during which no cows were being milked.
INPUT FORMAT
Line 1:The single integer, N
Lines 2~N+1:Two non-negative integers less than 1,000,000, respectively the starting and ending time in seconds after 0500
SAMPLE INPUT
3
300 1000
700 1200
1500 2100
OUTPUT FORMAT
A single line with two integers that represent the longest continuous time of milking and the longest idle time.
SAMPLE OUTPUT
900 300
C++编写：
这道题题意还是很简单，要求求一个最大挤奶时间(该时间段内至少有一头奶牛在挤奶)和最大间隔时间(该时间段内没有任何奶牛在挤奶) ，不过调试得我异常的艰辛，调了半天，结果就是一个地方写错了，后面会谈到，大家可以注意一下这些地方，那么代码如下：
/*
ID: your_id_here
TASK: milk2
LANG: C++
*/
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAX_N=5010;
struct Cow{
int begin;
int end;
}milk[MAX_N];
int N;
bool compare(const Cow &a,const Cow &b)
{
if(a.begin == b.begin)
return a.end<b.end;
return a.begin<b.begin; /*按照挤奶开始时间从早到晚排列，如果开始时间相同，则早结束的排在前面*/
}
void solve()
{
Cow t = milk[0];
int max_con=t.end-t.begin,max_idle=0;
int max1=max_con;
for(int i=1;i<N;i++)
{
if(milk[i].end <= t.end)
continue;
if(milk[i].begin <= t.end)
{
max1+=milk[i].end-t.end;
t=milk[i];
}
else
{
max_idle=max(max_idle,milk[i].begin-t.end);
t=milk[i];
max1=milk[i].end-milk[i].begin;
}
max_con=max(max_con,max1);
}
cout<<max_con<<" "<<max_idle<<endl;
}
int main()
{
freopen("milk2.in","r",stdin);
freopen("milk2.out","w",stdout);
cin>>N;
for(int i=0;i<N;i++)
cin>>milk[i].begin>>milk[i].end;
sort(milk,milk+N,compare);
solve();
return 0;
}
现在来谈一下调了很久才找到的错误，主要是solve()函数里面的for循环里有两行写得有问题，并且还不难发现，起初写出来的错误代码是下面这样的:
for(int i=1;i<N;i++)
{
if(milk[i].end <= t.end)
continue;
if(milk[i].begin <= t.end)
{
max1+=milk[i].end-milk[i-1].end;
t=milk[i];
}
else
{
max_idle=max(max_idle,milk[i].begin-milk[i-1].end);
t=milk[i];
max1=milk[i].end-milk[i].begin;
}
max_con=max(max_con,max1);
}
感觉上其实没问题，max1加上后一个的结束时间减去前一个的结束时间，当时USACO上1000个数据我也不想去找然后又不知道错哪儿，然后后来我自己想了两组数据，带入代码一看就明白了：
第一组(针对与求max_con会出现错误)：
4
5 10
6 8
7 9
9 12
理论应得：
7 0
实际会得：
8 0
第二组(针对于求max_idle会出现错误)：
4
5 10
6 8
7 9
12 15
理论应得：
5 2
实际会得：
5 3
